From: lrudolph@panix.com (Lee Rudolph)
Newsgroups: sci.math
Subject: Re: 1D knot in 4D?
Date: 11 Sep 1998 08:22:02 -0400

orourke@grendel.csc.smith.edu (Joseph O'Rourke) writes:

>Can anyone point me to a proof that no one-dimensional curve
>can be knotted in four-dimensional space?  Several knot theory
>textbooks discuss knotting 2-spheres in R^4, but I haven't
>found a proof at a 1-sphere cannot be knotted in R^4. Thanks!

Well, Charles Giffen seems to still be on vacation, so I'll take
a crack at this one.

First, you ought to decide what notion of "unknotted" you want
to use.  There are several potentially different ones (which
I *think* can all, with more or less effort, be shown to still
be the same in these dimensions; see footnote at the very end
for a cautionary tale).  I'll prove something, and let you deduce 
in what sense what I prove implies what you want.

Let C be a smoothly embedded simple closed curve in R^4.  The
space of all straight lines in R^4 is a real manifold V of 6 
dimensions (the space of all line segments in R^4 is 4+4=8 
dimensional, each straight line determines a subspace of
line segments of dimension 1+1=2, these subspaces are pairwise
disjoint, so dim V = 8-2=6).  

Let W be the subset of V consisting of (a) all lines which 
pass through 2 (or more) points of C, (b) all lines which 
are tangent to C (at at least one point).  Then W is the 
image by a smooth map of a smooth 2-manifold (namely, the 
symmetric square of C--the space of all unordered pairs of 
not necessarily distinct points of C: this space, homeomorphic
to the 2-sphere, is not a priori smooth along the "diagonal" 
copy of C, but it can be made smooth in a natural manner; and 
the assumed smoothness of C as a subset of R^4 is what you need
to show that the map from this space into V, onto W, is smooth).  

For any hyperplane H in R^4 (3-dimensional linear subspace),
let U(H) be the subset of V consisting of all lines perpendicular
to H.  Then U(H) is a 3-dimensional smooth submanifold of V.
Since dim(U(H))+dim(W)=3+2<6=dim V, "by transversality" a
generic choice of H will make the intersection of U(H) with W
empty.  By construction, then, orthogonal projection P from R^4
onto H will take C to a smoothly embedded curve C' in H, and
in fact you can find a 1-parameter family of diffeomorphisms
of R^4 (starting at the identity) which carries C onto C'.
So we've reduced the problem of "unknotting" C in R^4 to the 
case in which C = C' lies in R^3 contained in R^4.  And you
know how to do that already!  (Project C onto a generic 2-plane
in R^3 to get a "knot diagram" of the usual sort.  Choose
half, or fewer, of the crossings which, if changed, will
convert the diagram to a diagram of an unknot.  Move a small
part of the knot out into the 4th dimension to change those 
crossings.  Done.)

I think this proof is about as simple as you can get, 
unfortunately.  It has about the same level of difficulty
as an alternative proof using the "Whitney trick".  Both
rely on some sort of really "global" move (at the end).

Cautionary footnote: some sort of smoothness, or some restriction
on dimensions, *has* to be included in any general theorem about
unknottedness.  Consider the complex hypersurface X in C^4 defined
by the vanishing of x^2+y^3+z^5 (where (x,y,z,t) are the complex
coordinates in C^4, and you notice that t isn't used in the 
equation).  It's a fact that X (which has, of course, no natural
structure as a smooth manifold) is homeomorphic to R^6.  It's
also a fact that the t-axis, all points (0,0,0,t) in C^4, is
a subset T of X which is obviously homeomorphic to R^2.   
Yet this R^2 named T is so badly embedded in that R^6 named X
that its complement has non-trivial fundamental group (guess
which one [hint: 2, 3, 5!]).  Now THAT is knotted.

Lee Rudolph